Expand Using the Binomial Theorem (6y-11)^4

${(6 y - 11)}^{4}$

Step 1

Use the binomial expansion theorem to find each term. The binomial theorem states ${(a + b)}^{n} = \sum_{k = 0}^{n} n C k \cdot (a^{n - k} b^{k})$ .

$\sum_{k = 0}^{4} \frac{4!}{(4 - k)! k!} \cdot {(6 y)}^{4 - k} \cdot {(- 11)}^{k}$

Step 2

Expand the summation.

$\frac{4!}{(4 - 0)! 0!} \cdot {(6 y)}^{4 - 0} \cdot {(- 11)}^{0} + \frac{4!}{(4 - 1)! 1!} \cdot {(6 y)}^{4 - 1} \cdot {(- 11)}^{1} + \frac{4!}{(4 - 2)! 2!} \cdot {(6 y)}^{4 - 2} \cdot {(- 11)}^{2} + \frac{4!}{(4 - 3)! 3!} \cdot {(6 y)}^{4 - 3} \cdot {(- 11)}^{3} + \frac{4!}{(4 - 4)! 4!} \cdot {(6 y)}^{4 - 4} \cdot {(- 11)}^{4}$

Step 3

Simplify the exponents for each term of the expansion.

$1 \cdot {(6 y)}^{4} \cdot {(- 11)}^{0} + 4 \cdot {(6 y)}^{3} \cdot {(- 11)}^{1} + 6 \cdot {(6 y)}^{2} \cdot {(- 11)}^{2} + 4 \cdot {(6 y)}^{1} \cdot {(- 11)}^{3} + 1 \cdot {(6 y)}^{0} \cdot {(- 11)}^{4}$

Step 4

Simplify each term.

Multiply ${(6 y)}^{4}$ by $1$ .

${(6 y)}^{4} \cdot {(- 11)}^{0} + 4 \cdot {(6 y)}^{3} \cdot {(- 11)}^{1} + 6 \cdot {(6 y)}^{2} \cdot {(- 11)}^{2} + 4 \cdot {(6 y)}^{1} \cdot {(- 11)}^{3} + 1 \cdot {(6 y)}^{0} \cdot {(- 11)}^{4}$

Apply the product rule to $6 y$ .

$6^{4} y^{4} \cdot {(- 11)}^{0} + 4 \cdot {(6 y)}^{3} \cdot {(- 11)}^{1} + 6 \cdot {(6 y)}^{2} \cdot {(- 11)}^{2} + 4 \cdot {(6 y)}^{1} \cdot {(- 11)}^{3} + 1 \cdot {(6 y)}^{0} \cdot {(- 11)}^{4}$

Raise $6$ to the power of $4$ .

$1296 y^{4} \cdot {(- 11)}^{0} + 4 \cdot {(6 y)}^{3} \cdot {(- 11)}^{1} + 6 \cdot {(6 y)}^{2} \cdot {(- 11)}^{2} + 4 \cdot {(6 y)}^{1} \cdot {(- 11)}^{3} + 1 \cdot {(6 y)}^{0} \cdot {(- 11)}^{4}$

Anything raised to $0$ is $1$ .

$1296 y^{4} \cdot 1 + 4 \cdot {(6 y)}^{3} \cdot {(- 11)}^{1} + 6 \cdot {(6 y)}^{2} \cdot {(- 11)}^{2} + 4 \cdot {(6 y)}^{1} \cdot {(- 11)}^{3} + 1 \cdot {(6 y)}^{0} \cdot {(- 11)}^{4}$

Multiply $1296$ by $1$ .

$1296 y^{4} + 4 \cdot {(6 y)}^{3} \cdot {(- 11)}^{1} + 6 \cdot {(6 y)}^{2} \cdot {(- 11)}^{2} + 4 \cdot {(6 y)}^{1} \cdot {(- 11)}^{3} + 1 \cdot {(6 y)}^{0} \cdot {(- 11)}^{4}$

Apply the product rule to $6 y$ .

$1296 y^{4} + 4 \cdot (6^{3} y^{3}) \cdot {(- 11)}^{1} + 6 \cdot {(6 y)}^{2} \cdot {(- 11)}^{2} + 4 \cdot {(6 y)}^{1} \cdot {(- 11)}^{3} + 1 \cdot {(6 y)}^{0} \cdot {(- 11)}^{4}$

Raise $6$ to the power of $3$ .

$1296 y^{4} + 4 \cdot (216 y^{3}) \cdot {(- 11)}^{1} + 6 \cdot {(6 y)}^{2} \cdot {(- 11)}^{2} + 4 \cdot {(6 y)}^{1} \cdot {(- 11)}^{3} + 1 \cdot {(6 y)}^{0} \cdot {(- 11)}^{4}$

Multiply $216$ by $4$ .

$1296 y^{4} + 864 y^{3} \cdot {(- 11)}^{1} + 6 \cdot {(6 y)}^{2} \cdot {(- 11)}^{2} + 4 \cdot {(6 y)}^{1} \cdot {(- 11)}^{3} + 1 \cdot {(6 y)}^{0} \cdot {(- 11)}^{4}$

Evaluate the exponent.

$1296 y^{4} + 864 y^{3} \cdot - 11 + 6 \cdot {(6 y)}^{2} \cdot {(- 11)}^{2} + 4 \cdot {(6 y)}^{1} \cdot {(- 11)}^{3} + 1 \cdot {(6 y)}^{0} \cdot {(- 11)}^{4}$

Multiply $- 11$ by $864$ .

$1296 y^{4} - 9504 y^{3} + 6 \cdot {(6 y)}^{2} \cdot {(- 11)}^{2} + 4 \cdot {(6 y)}^{1} \cdot {(- 11)}^{3} + 1 \cdot {(6 y)}^{0} \cdot {(- 11)}^{4}$

Apply the product rule to $6 y$ .

$1296 y^{4} - 9504 y^{3} + 6 \cdot (6^{2} y^{2}) \cdot {(- 11)}^{2} + 4 \cdot {(6 y)}^{1} \cdot {(- 11)}^{3} + 1 \cdot {(6 y)}^{0} \cdot {(- 11)}^{4}$

Multiply $6$ by $6^{2}$ by adding the exponents.

Move $6^{2}$ .

$1296 y^{4} - 9504 y^{3} + 6^{2} \cdot 6 \cdot y^{2} \cdot {(- 11)}^{2} + 4 \cdot {(6 y)}^{1} \cdot {(- 11)}^{3} + 1 \cdot {(6 y)}^{0} \cdot {(- 11)}^{4}$

Multiply $6^{2}$ by $6$ .

Raise $6$ to the power of $1$ .

$1296 y^{4} - 9504 y^{3} + 6^{2} \cdot 6^{1} \cdot y^{2} \cdot {(- 11)}^{2} + 4 \cdot {(6 y)}^{1} \cdot {(- 11)}^{3} + 1 \cdot {(6 y)}^{0} \cdot {(- 11)}^{4}$

Use the power rule $a^{m} a^{n} = a^{m + n}$ to combine exponents.

$1296 y^{4} - 9504 y^{3} + 6^{2 + 1} \cdot y^{2} \cdot {(- 11)}^{2} + 4 \cdot {(6 y)}^{1} \cdot {(- 11)}^{3} + 1 \cdot {(6 y)}^{0} \cdot {(- 11)}^{4}$

Add $2$ and $1$ .

$1296 y^{4} - 9504 y^{3} + 6^{3} \cdot y^{2} \cdot {(- 11)}^{2} + 4 \cdot {(6 y)}^{1} \cdot {(- 11)}^{3} + 1 \cdot {(6 y)}^{0} \cdot {(- 11)}^{4}$

Raise $6$ to the power of $3$ .

$1296 y^{4} - 9504 y^{3} + 216 \cdot y^{2} \cdot {(- 11)}^{2} + 4 \cdot {(6 y)}^{1} \cdot {(- 11)}^{3} + 1 \cdot {(6 y)}^{0} \cdot {(- 11)}^{4}$

Raise $- 11$ to the power of $2$ .

$1296 y^{4} - 9504 y^{3} + 216 y^{2} \cdot 121 + 4 \cdot {(6 y)}^{1} \cdot {(- 11)}^{3} + 1 \cdot {(6 y)}^{0} \cdot {(- 11)}^{4}$

Multiply $121$ by $216$ .

$1296 y^{4} - 9504 y^{3} + 26136 y^{2} + 4 \cdot {(6 y)}^{1} \cdot {(- 11)}^{3} + 1 \cdot {(6 y)}^{0} \cdot {(- 11)}^{4}$

Simplify.

$1296 y^{4} - 9504 y^{3} + 26136 y^{2} + 4 \cdot (6 y) \cdot {(- 11)}^{3} + 1 \cdot {(6 y)}^{0} \cdot {(- 11)}^{4}$

Multiply $6$ by $4$ .

$1296 y^{4} - 9504 y^{3} + 26136 y^{2} + 24 y \cdot {(- 11)}^{3} + 1 \cdot {(6 y)}^{0} \cdot {(- 11)}^{4}$

Raise $- 11$ to the power of $3$ .

$1296 y^{4} - 9504 y^{3} + 26136 y^{2} + 24 y \cdot - 1331 + 1 \cdot {(6 y)}^{0} \cdot {(- 11)}^{4}$

Multiply $- 1331$ by $24$ .

$1296 y^{4} - 9504 y^{3} + 26136 y^{2} - 31944 y + 1 \cdot {(6 y)}^{0} \cdot {(- 11)}^{4}$

Multiply ${(6 y)}^{0}$ by $1$ .

$1296 y^{4} - 9504 y^{3} + 26136 y^{2} - 31944 y + {(6 y)}^{0} \cdot {(- 11)}^{4}$

Apply the product rule to $6 y$ .

$1296 y^{4} - 9504 y^{3} + 26136 y^{2} - 31944 y + 6^{0} y^{0} \cdot {(- 11)}^{4}$

Anything raised to $0$ is $1$ .

$1296 y^{4} - 9504 y^{3} + 26136 y^{2} - 31944 y + 1 y^{0} \cdot {(- 11)}^{4}$

Multiply $y^{0}$ by $1$ .

$1296 y^{4} - 9504 y^{3} + 26136 y^{2} - 31944 y + y^{0} \cdot {(- 11)}^{4}$

Anything raised to $0$ is $1$ .

$1296 y^{4} - 9504 y^{3} + 26136 y^{2} - 31944 y + 1 \cdot {(- 11)}^{4}$

Multiply ${(- 11)}^{4}$ by $1$ .

$1296 y^{4} - 9504 y^{3} + 26136 y^{2} - 31944 y + {(- 11)}^{4}$

Raise $- 11$ to the power of $4$ .

$1296 y^{4} - 9504 y^{3} + 26136 y^{2} - 31944 y + 14641$

Expand Using the Binomial Theorem (6y-11)^4

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